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airyana1114

  • one year ago

Can someone help me solve this equation? Step by Step.

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    is it a hard equation ?

  3. airyana1114
    • one year ago
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    \[\sqrt{x+6}-4=x\]

  4. misty1212
    • one year ago
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    oh no that can't be that hard

  5. airyana1114
    • one year ago
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    Can you help me? I've never had this type of equation equaled to x and it sort of makes me nervous. Can you walk me through it? @misty1212

  6. misty1212
    • one year ago
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    yeah no problem first step is to get the radical by itself on one side of the equation by adding 4 to both sides, i.e. write \[\sqrt{x+6}=x+4\]

  7. misty1212
    • one year ago
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    now you want to do it the math teacher way , or you want to think what the answer is , or both?

  8. airyana1114
    • one year ago
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    I'm not sure I understand what you're asking.

  9. misty1212
    • one year ago
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    we can find the answer in our head, but that is probably not what your math teacher wants you to do ,so lets go ahead and solve it the way they do in math class

  10. airyana1114
    • one year ago
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    Okay

  11. misty1212
    • one year ago
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    i can show you how to get the answer in your head second

  12. airyana1114
    • one year ago
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    Alright thanks :)

  13. misty1212
    • one year ago
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    you want to get rid of the radical, so square both sides (carefully)

  14. airyana1114
    • one year ago
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    How would I square (x+4). Like this? x^2+8

  15. misty1212
    • one year ago
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    \[\sqrt{x+6}=x+4\] square and get \[x+6=(x+4)^2\] or \[x+6=(x+4)(x+4)\] or \[x+6=x^2+8x+16\]

  16. airyana1114
    • one year ago
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    Where did the 8x come from?

  17. misty1212
    • one year ago
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    no dear that is not how you square \(x+4\) you have to write it as \[x(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16\]

  18. misty1212
    • one year ago
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    it is that old "first outer inner last" stuff

  19. airyana1114
    • one year ago
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    Ohhhh, FOIL okay..that makes sense.

  20. misty1212
    • one year ago
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    so now we have \[x+6=x^2+8x+16\] which is a quadratic equation we have to solve

  21. misty1212
    • one year ago
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    set it equal to zero, get \[x^2+7x+10=0\] then factor

  22. misty1212
    • one year ago
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    \[x^2+7x+10=0\\ (x+2)(x+5)=0\] so \[x=-2,x=-5\] but we are not done yet

  23. misty1212
    • one year ago
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    you have to check the answer in the original equation not the quadratic equation if you do, you will see that \(x=-2\) works but \(x=-5\) does not

  24. airyana1114
    • one year ago
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    Making it an extraneous solution correct?

  25. misty1212
    • one year ago
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    right , which is a silly way of saying it is not a solution but no matter

  26. airyana1114
    • one year ago
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    :) thank you @misty1212

  27. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

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